Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 years ago

Teacher gave my group 13 problems to solve, and we can't solve this one. So please help.

Problem 6. The exponential of a square matrix M is, by definition, the

limit of the following sequence

N

e^M = 1 + M + M2/2! + · · · = Lim ∑ M^k / k!

N→+∞ k=0

.

One admits that this limit exists, by a theorem of Analysis.

1. Show that if AB = BA then e^(A+B) = e^A e^B One is allowed, to treat

this question, to pass to the limit without justification.

2. Compute e^M for the four following matrices :

(a 0 0)

(0 b 0)

(0 0 c)

(0 a b)

(0 0 c)

(0 0 0)

(0 1)

(−1 0 )

(1 0 )

(0 0 )

3. Find a simple example where e^(A+B) does not equal e^A e^B

Relevance
• kb
Lv 7
2 years ago

1) The crucial fact is that the Binomial Theorem extends to square matrices if AB = BA:

If AB = BA, then (A + B)^k = Σ(j = 0 to k) C(k, j) A^j B^(k-j).

For instance, look at the case when k = 2.

(A + B)^2 = (A + B)(A + B)

................= A(A + B) + B(A + B)

................= A^2 + AB + BA + B^2

................= A^2 + 2AB + B^2, only if we assume that AB = BA.

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From here, this question is not too difficult.

e^(A+B) = Σ(k = 0 to ∞) (A+B)^k / k!

..............= Σ(k = 0 to ∞) Σ(j = 0 to k) C(k, j) A^j B^(k-j) / k!, by above

..............= Σ(k = 0 to ∞) Σ(j = 0 to k) (k!/(j! (k-j)!)) * A^j B^(k-j) / k!

..............= Σ(k = 0 to ∞) Σ(j = 0 to k) A^j B^(k-j) / (j! (k-j)!)

..............= Σ(j = 0 to ∞) Σ(n = 0 to ∞) A^j B^n / (j! n!), letting k = j+n

..............= Σ(j = 0 to ∞) A^j/j! * Σ(n = 0 to ∞) B^n/n!

..............= e^A e^B.

----

2) Since D =

(a 0 0)

(0 b 0)

(0 0 c) is a diagonal matrix, we have that

D^n =

(a^n 0 0)

(0 b^n 0)

(0 0 c^n) for any n = 0, 1, 2, ... .

Hence, e^D

= Σ(n = 0 to ∞) D^n / n!

= (Σ(n = 0 to ∞) a^n/n! 0 0)

...(0 Σ(n = 0 to ∞) b^n/n! 0)

...(0 0 Σ(n = 0 to ∞) c^n/n!), which equals

(e^a 0 0)

(0 e^b 0)

(0 0 e^c).

---

ii) Let A =

(0 a b)

(0 0 c)

(0 0 0).

So, A^2 =

(0 0 ac)

(0 0 0)

(0 0 0), and A^n = 0 for all n = 3, 4, 5, ...

So, e^A = I + A + (1/2!)A^2 + 0 =

(1 a b+(ac/2))

(0 1 c)

(0 0 1).

---

iii) Let B =

(0 1)

(-1 0).

Check that B^2 = -I, B^3 = -B, and B^4 = I.

(This means that the powers of B go in cycles of 4.)

So, e^B

= I + B + B^2/2! + B^3/3! + B^4/4! + B^5/5! + B^6/6! + B^7/7! + B^8/8! + ...

= I + B - I/2! - B/3! + I/4! + B/5! - I/6! - B/7! + I/8! + ...

= (1 - 1/2! + 1/4! - 1/6! + 1/8! - ...)I + (1 - 1/3! + 1/5! - 1/7! + ...)B

= (cos 1)I + (sin 1)B

= [cos 1 sin 1]

...[sin 1 cos 1].

---

iv) This is similar to (i), and you obtain

(e^1 0)

(0 e^0).

----

3) As long as you choose A and B so that AB does not equal BA, you should be fine.

For example, let A = (1 0; 0 0) and B = (0 0; 0 1).

Then, as above, we have

e^A = (e 0; 0 1)

e^B = I + B + 0, noting that B^2 = 0

.......= (1 1; 0 1)

and e^(A+B) = e^I = (e 0; 0 e).

However, e^(A+B) does not equal e^A e^B.

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I hope this helps!

• Cheatawin2 years agoReport

You're the best, man. :'D